Oscillations — SHM, Energy & Pendulum for Class 11

What Actually Happens in Oscillations

Take a pendulum. Push it, and it swings back. Push a spring, it pushes back. What’s common? The system always tries to return to where it started — and that “trying to return” is the entire story of oscillations.

The force responsible is called the restoring force, and it always points toward the mean position (equilibrium). When this restoring force is directly proportional to displacement — and opposite in direction — we get Simple Harmonic Motion (SHM). That one condition gives us an enormous amount of structure to work with.

Formally: $F = -kx$ , where $k$ is a positive constant and $x$ is displacement from mean position. The negative sign is doing all the work here — it ensures the force opposes displacement.

SHM isn’t just a textbook abstraction. Every vibrating string, every AC circuit, every molecule in a crystal lattice, every bridge that can resonate — they all trace back to this same mathematics. Class 11 is where you build the foundation; JEE builds the walls.

Key Terms & Definitions

Mean Position (Equilibrium): The point where net force on the particle is zero. In SHM, this is also where speed is maximum.

Amplitude (A): Maximum displacement from mean position. Units: metres. Amplitude tells you how far the particle goes — it does not affect time period (crucial for JEE).

Time Period (T): Time taken to complete one full oscillation. Units: seconds.

Frequency (f or ν): Number of oscillations per second. $f = \frac{1}{T}$ . Units: Hz (hertz).

Angular Frequency (ω): $\omega = 2\pi f = \frac{2\pi}{T}$ . Units: rad/s. Most SHM equations are cleaner in terms of ω.

Phase (φ): Tells you where in its cycle the particle is at $t = 0$ . The general displacement equation is:

x(t) = A\sin(\omega t + \phi)

or equivalently $A\cos(\omega t + \phi)$ — both are valid, choose based on initial conditions.

Phase Difference: Between two SHM particles. If phase difference is $\pi$ , they move in exactly opposite directions at every instant.

The Core Equations of SHM

x = A\sin(\omega t + \phi)

v = A\omega\cos(\omega t + \phi) = \omega\sqrt{A^2 - x^2}

a = -A\omega^2\sin(\omega t + \phi) = -\omega^2 x

T = \frac{2\pi}{\omega}

The velocity equation $v = \omega\sqrt{A^2 - x^2}$ is the one you’ll use most in problems. Memorise the shape: velocity is maximum at $x = 0$ (mean position) and zero at $x = \pm A$ (extreme positions).

The acceleration equation $a = -\omega^2 x$ is the defining condition for SHM. If you can write acceleration as proportional to $-x$ , you’ve confirmed SHM and extracted ω in one step.

Deriving ω from the Equation of Motion

This is the standard technique for any SHM problem:

Write the equation of motion: $ma = F_{net}$
Express $a$ in terms of displacement $x$
Compare with $a = -\omega^2 x$ to read off $\omega$
Then $T = \frac{2\pi}{\omega}$

We’ll apply this to both spring-mass and pendulum systems below.

Spring-Mass System

A block of mass $m$ on a frictionless surface, attached to a spring of spring constant $k$ .

When displaced by $x$ : restoring force $F = -kx$

So: $ma = -kx \Rightarrow a = -\frac{k}{m}x$

Comparing with $a = -\omega^2 x$ :

\omega = \sqrt{\frac{k}{m}}

T = 2\pi\sqrt{\frac{m}{k}} \qquad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

Notice T depends on $\sqrt{m}$ and $\sqrt{1/k}$ . If you double the mass, T increases by factor $\sqrt{2}$ . If you double k (stiffer spring), T decreases — stiffer spring pulls back harder, oscillates faster.

Springs in Series and Parallel — a JEE favourite:

Two springs in parallel: $k_{eff} = k_1 + k_2$
Two springs in series: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$

Parallel springs feel like one strong spring. Series springs feel like one weak spring. Same logic as resistors, but the parallel-series roles are swapped compared to what you might expect.

Spring cut in ratio 1:n: If a spring of constant $k$ and length $L$ is cut into lengths $\ell$ and $L - \ell$ , the spring constants become $k\frac{L}{\ell}$ and $k\frac{L}{L-\ell}$ . Shorter piece = stiffer spring.

Simple Pendulum

A point mass $m$ suspended by a massless string of length $L$ , displaced by small angle $\theta$ .

The restoring force along the arc is $-mg\sin\theta$ . For small angles ( $\theta < 15°$ ), $\sin\theta \approx \theta$ , so:

F = -mg\theta = -mg\frac{x}{L}

This is now in the form $F = -kx$ with effective $k = \frac{mg}{L}$ .

T = 2\pi\sqrt{\frac{L}{g}} \qquad \omega = \sqrt{\frac{g}{L}}

Key results worth memorising:

T is independent of mass and amplitude (for small oscillations)
T depends only on L and g
On the Moon: $g_{moon} = g_{earth}/6$ , so $T_{moon} = \sqrt{6} \cdot T_{earth}$ — pendulum swings slower

Effective gravity situations (JEE loves these):

Situation	Effective g	Effect on T
Lift accelerating up	$g + a$	T decreases
Lift accelerating down	$g - a$	T increases
Lift in free fall	0	T → ∞ (pendulum won’t oscillate)
Charged pendulum in electric field	$g \pm \frac{qE}{m}$	Depends on direction

JEE Main pattern: Problems where a pendulum is in an accelerating frame, or where temperature change alters the string length. For thermal expansion: $\frac{\Delta T}{T} = \frac{1}{2}\alpha\Delta\theta$ , where $\alpha$ is the coefficient of linear expansion. This appeared in JEE Main 2023 Shift 2.

Energy in SHM

At any displacement $x$ :

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

PE = \frac{1}{2}m\omega^2 x^2

E_{total} = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2

Total energy is constant — it just shifts between KE and PE. At mean position: all KE, zero PE. At extremes: all PE, zero KE.

KE = PE condition:

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 x^2

A^2 - x^2 = x^2 \Rightarrow x = \pm\frac{A}{\sqrt{2}}

So KE = PE at $x = \frac{A}{\sqrt{2}}$ from mean position. This is a standard result — the particle spends equal KE and PE at roughly 70.7% of its amplitude.

Energy-time graphs: Both KE and PE vary as $\sin^2$ or $\cos^2$ , so they oscillate at twice the frequency of displacement. Total energy is a horizontal line. This graph question is a 1-mark CBSE staple.

Solved Examples

Example 1 — Easy (CBSE Level)

A particle executes SHM with amplitude 5 cm and time period 0.2 s. Find the maximum velocity and maximum acceleration.

$A = 5 \times 10^{-2}$ m, $T = 0.2$ s

\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi \text{ rad/s}

v_{max} = A\omega = 5 \times 10^{-2} \times 10\pi = 0.5\pi \approx 1.57 \text{ m/s}

a_{max} = A\omega^2 = 5 \times 10^{-2} \times (10\pi)^2 = 5\pi^2 \approx 49.3 \text{ m/s}^2

Example 2 — Medium (JEE Main Level)

A spring-mass system has T = 2s on Earth. What is T when the same system is taken to the Moon?

Time period of spring-mass: $T = 2\pi\sqrt{\frac{m}{k}}$

This has no g dependence. So T remains 2s on the Moon.

Many students reflexively change T when the problem says “Moon.” Only the simple pendulum depends on g — the spring-mass system does not. Know which formula contains g and which doesn’t.

Example 3 — Hard (JEE Main/Advanced Level)

Two blocks $m_1 = 1$ kg and $m_2 = 2$ kg are connected by a spring (k = 300 N/m). They are placed on a frictionless surface and given initial compression of 5 cm. Find the time period of oscillation.

When two masses are connected by a spring on a frictionless surface, both oscillate. The system is equivalent to a single mass equal to the reduced mass:

\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \text{ kg}

T = 2\pi\sqrt{\frac{\mu}{k}} = 2\pi\sqrt{\frac{2/3}{300}} = 2\pi\sqrt{\frac{1}{450}} = \frac{2\pi}{15\sqrt{2}} \approx 0.296 \text{ s}

The reduced mass concept is often skipped in NCERT but appears consistently in JEE Main.

Exam-Specific Tips

CBSE Class 11 Board Exam

Chapter 14 typically carries 5-8 marks in theory + numericals.
Derivation of $T = 2\pi\sqrt{L/g}$ for simple pendulum is asked almost every year — practice writing it cleanly with the small-angle approximation step explicitly shown.
Energy diagrams (KE vs time, PE vs time, total E vs time) appear as short-answer questions.
For 3-mark numericals: show all substitutions explicitly, the marking scheme awards 1 mark per step.

JEE Main

Oscillations carries 3-4 questions on average (weightage: ~8-10%).
Focus areas: spring combinations, pendulum in non-inertial frames, superposition of two SHMs, energy-based questions.
The “find x when KE = nPE” type is a 1-mark quick question — learn the formula $x = A\sqrt{\frac{n}{n+1}}$ .

JEE Advanced occasionally tests coupled oscillations and damped/forced oscillations conceptually. For Class 11 board prep, stick to undamped free oscillations. But if you’re targeting advanced, read about resonance conditions — the concept of natural frequency vs driving frequency.

SAT Physics / International

SAT Subject Test (legacy): focus on period-frequency relationship, energy conservation in spring.
AP Physics C: uses calculus — SHM is derived from $F = -kx$ using differential equations. Good to know if you’re targeting US colleges.

Common Mistakes to Avoid

Mistake 1: Confusing x with A in velocity formula. Students write $v = \omega\sqrt{A^2 - A^2}$ when asked for velocity at extreme — and get zero (correct!), but then write $v = \omega\sqrt{A^2}$ thinking x = 0 at mean position. Make sure you know: x = 0 at mean position, x = ±A at extreme positions.

Mistake 2: Applying T = 2π√(L/g) to spring-mass. T for spring-mass is $2\pi\sqrt{m/k}$ . These are different systems. The pendulum formula has g; the spring formula has k. Write them side by side once and the difference becomes permanent.

Mistake 3: Forgetting the negative sign in a = −ω²x. When checking if a motion is SHM, write a = −(some positive constant) × x. If your constant comes out negative, you’ve made a sign error somewhere. SHM requires that constant to be positive.

Mistake 4: Phase confusion — sin vs cos. $x = A\sin\omega t$ means particle starts at mean position (x = 0 at t = 0). $x = A\cos\omega t$ means particle starts at extreme (x = A at t = 0). Read the initial condition of the problem carefully before choosing your form.

Mistake 5: Wrong spring combination formula. Parallel springs: k adds (like conductances). Series springs: 1/k adds (like resistances). Students often reverse these. Memory trick: parallel springs both stretch/compress the same amount = same as a stiffer spring = k adds up.

Practice Questions

Q1. A particle in SHM has displacement $x = 4\sin(2\pi t + \pi/6)$ cm. Find: (a) amplitude, (b) time period, (c) initial displacement.

(a) A = 4 cm
(b) ω = 2π rad/s, so T = 2π/ω = 1 s
(c) At t = 0: x = 4sin(π/6) = 4 × 0.5 = 2 cm

Q2. A spring of spring constant 400 N/m has a block of 100 g attached. Find the time period and frequency of oscillation.

m = 0.1 kg, k = 400 N/m
$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.1/400} = 2\pi\sqrt{1/4000} = 2\pi/(20\sqrt{10}) \approx 0.099$ s
$f = 1/T \approx 10.1$ Hz

Q3. Find the length of a simple pendulum with T = 2 s on Earth (g = 9.8 m/s²). This is the “seconds pendulum.”

$T = 2\pi\sqrt{L/g}$
$2 = 2\pi\sqrt{L/9.8}$
$1 = \pi\sqrt{L/9.8}$
$L/9.8 = 1/\pi^2$
$L = 9.8/\pi^2 \approx 0.993$ m ≈ 1 m
The seconds pendulum is approximately 1 metre long.

Q4. In SHM, the displacement is $x = A\sin\omega t$ . At what displacement is KE = 3 × PE?

KE = 3PE
$\frac{1}{2}m\omega^2(A^2 - x^2) = 3 \times \frac{1}{2}m\omega^2 x^2$
$A^2 - x^2 = 3x^2$
$A^2 = 4x^2$
$x = \pm\frac{A}{2}$

Q5. A spring of natural length 30 cm is cut into two pieces of lengths 10 cm and 20 cm. If original spring constant is 60 N/m, find the spring constants of both pieces.

Spring constant × length = constant (for a uniform spring)
$k_1 \times 10 = 60 \times 30 \Rightarrow k_1 = 180$ N/m
$k_2 \times 20 = 60 \times 30 \Rightarrow k_2 = 90$ N/m
Shorter piece is stiffer.

Q6. A particle executes SHM. At displacement x = 3 cm, its velocity is 4 cm/s. At displacement x = 4 cm, its velocity is 3 cm/s. Find amplitude.

Using $v = \omega\sqrt{A^2 - x^2}$ :
$16 = \omega^2(A^2 - 9)$ … (i)
$9 = \omega^2(A^2 - 16)$ … (ii)
Dividing: $\frac{16}{9} = \frac{A^2 - 9}{A^2 - 16}$
$16(A^2 - 16) = 9(A^2 - 9)$
$16A^2 - 256 = 9A^2 - 81$
$7A^2 = 175$
$A^2 = 25 \Rightarrow A = 5$ cm

Q7. The time period of a pendulum on Earth is 2 s. A lift accelerates downward at $g/4$ . Find T in the lift.

Effective g = g - g/4 = 3g/4
$T_{lift} = 2\pi\sqrt{L/(3g/4)} = 2\pi\sqrt{4L/3g}$
$= \frac{2}{\sqrt{3/4}} \times 2\pi\sqrt{L/g} \times \frac{2}{\sqrt{3}}$
Since original T = 2 s: $T_{lift} = T\sqrt{4/3} = 2 \times \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}} \approx 2.31$ s
(Pendulum slows down — effective gravity is weaker.)

Q8. Total energy in SHM is 0.5 J. When displacement is half the amplitude, what is the KE?

$E_{total} = 0.5$ J
At $x = A/2$ :
$PE = \frac{1}{2}k(A/2)^2 = \frac{1}{4} \times \frac{1}{2}kA^2 = \frac{E_{total}}{4} = 0.125$ J
$KE = E_{total} - PE = 0.5 - 0.125 = 0.375$ J
KE is three-fourths of total energy at $x = A/2$ .

FAQs

Why is the time period of a simple pendulum independent of mass?

Because the restoring force is gravity, and gravity accelerates all masses equally ( $F = mg$ , $a = g$ ). The mass cancels out when you write the equation of motion. This is the same reason all objects fall at the same rate in vacuum — Galileo’s experiment in equation form.

What is the difference between oscillation and vibration?

Technically, they’re used interchangeably in Class 11. In engineering, “vibration” often refers to mechanical systems and “oscillation” to electrical or general periodic motion. For CBSE and JEE, treat them as synonyms.

Can SHM occur without a restoring force?

No. SHM is defined by $F = -kx$ . Without a restoring force proportional to displacement, the motion may be periodic but it won’t be SHM. Circular motion is periodic but not SHM because the force direction is different.

How does damping affect time period?

In damped SHM, the time period slightly increases and amplitude decreases exponentially. For Class 11 boards, this is conceptual only — you won’t need to calculate it. JEE occasionally asks conceptual questions about underdamped vs overdamped systems.

Why does SHM need small-angle approximation for a pendulum?

Because $\sin\theta \approx \theta$ only holds for small $\theta$ (say, less than 15°). For larger angles, the restoring force is $-mg\sin\theta$ , which is not proportional to $\theta$ — so it’s no longer SHM. The motion is still periodic, but the time period depends on amplitude. This is why we say “for small oscillations” every time.

What happens to the spring-mass system in a satellite (weightlessness)?

The spring-mass T depends only on $m/k$ — no g involved. So it oscillates normally even in weightlessness. A pendulum in a satellite won’t oscillate at all since effective g = 0.

How is SHM related to uniform circular motion?

SHM is the projection of uniform circular motion onto a diameter. If a particle moves in a circle of radius A with angular velocity ω, its shadow on the diameter executes SHM with amplitude A and angular frequency ω. This geometric connection makes the equations natural — $x = A\cos\omega t$ directly corresponds to the x-coordinate of circular motion.

What is the significance of ω²x = acceleration in checking SHM?

It’s the fastest way to identify and characterize SHM in any system. Write the net force, convert to acceleration, factor out the displacement term. Whatever multiplies $x$ (with a negative sign) equals $\omega^2$ . From ω, you get T, f, and all other characteristics immediately.